Question

What is the value of x? Enter your answer in the box.

Answer 1

The value of `color(blue)(x~~78.91 feet` (2 decimal places)

Explanation:

The question has a note that reads Image not drawn to scale.

I am assuming that the Line segment ED is the Altitude of the triangle.

The altitude of the triangle GEH is a Perpendicular line drawn from the Vertex E on to the opposite side GH.

Hence, the altitude ED may be referred as a line segment and forms the Right angle with the edge GH opposite to the vertex E.

We can conclude that the triangles GDE and HDE are both Right triangles.

Hence, angle at `D` is 90 degrees for both the triangles GDE and HDE.

Pythagorean Theorem states that:

for all right-angled triangles, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

The hypotenuse is the longest side and it's always opposite the right angle.

Hence, in our triangle GDE,

`GE^2=GD^2 + DE^2`

We are given that

`Side " GD" = 62` feet and `Side " GE"=99.2` feet

Hence,

`(99.2)^2 = (62)^2 + DE^2`

Subtract `(62)^2` from both sides to simplify.

`(99.2)^2 - (62)^2 = (62)^2 + DE^2- (62)^2`

`(99.2)^2 - (62)^2 = cancel (62)^2 + DE^2- cancel (62)^2`

`rArr DE^2 = (99.2)^2 - (62)^2`

`rArr DE^2 = 9840.64 - 3844`

`rArr DE^2 = 5996.64`

`rArr DE = sqrt(5996.64)`

`rArr DE ~~ 77.44" "`feet

Next, consider the right-triangle HDE.

Using the Pythagorean Theorem, we can write

`HE^2=HD^2 + DE^2`

`rArr (112)^2 = (x+2)^2 + (77.44)^2`

Subtract `(77.44)^2` from both sides.

`rArr (112)^2 - (77.44)^2 = (x+2)^2 + (77.44)^2 - (77.44)^2`

`rArr (112)^2 - (77.44)^2 = (x+2)^2 + cancel (77.44)^2 - cancel (77.44)^2`

`rArr (112)^2 - (77.44)^2 = (x+2)^2`

`rArr 12544 - 5996.954 = (x+2)^2`

`rArr 6547.046 = (x+2)^2`

`rArr (x+2)^2 = 6547.046`

`rArr (x+2) = sqrt(6547.046)`

`rArr (x+2) ~~ 80.91382`

Subtract `2` from both sides.

`rArr x+2-2 ~~ 80.91382-2`

`rArr x+cancel 2- cancel 2 ~~ 80.91382-2`

`rArr x ~~ 80.91382-2`

`rArr x ~~ 78.91382`

The value of `color(blue)(x~~78.91 " feet"` (2 decimal places)

Answer 2

Case 1 : ED is the altitude (perpendicular to GH) `x = color(brown)(78.92` ft

Case 2 : ED is the median `x = color(blue)(60` ft

Case 3 :ED is the angular bisector of `GhatEH` `x = color(green)(68` ft

Explanation:

Assuming

Since it has not been given, ED is perpendicular to GH or D is the mid point of GH or ED is the angular bisector of `GhatEH`

Let's consider all these three cases.

Case 1 : ED is the altitude (perpendicular to GH)

In right triangle DEG,

`DE^2 = GE62 - DG^2 = 99.2^2 - 62^2` Eqn (1)

In right triangle DEH,

`DE^2 = EH^2 - DH^2 = 112^2 - (x+2)^2` Eqn (2)

Comparing Eqns (1), (2),

`112^2 - (x+2)^2 = 99.2^2 - 62^2`

`(x+2)^2 = 112^2 - 99.2^2 + 62^2 = 6547.36`

`(x+2) ~~ 80.92`

`x = 80.92 - 2 = color(brown)(78.92` ft

Case 2 : ED is the median

i.e DH = DG #

`x + 2 = 62` or `x = 62 - 2 = color(blue)(60` ft

Case 3 : ED is the angular bisector of #GhatEH

As per angular bisector theorem,

`(EG) / (EH) = (DG) / (GH)`

`99.2 / 112 = 62 / (x + 2)`

`(x +2) = (62 * 112) / 99.2 = 70`

`x = 70 - 2 = color(green)(68)` ft